Question

Which of the following contains less number of molecules?

• A. $11gm$ of $C{O}_2$
• B. $32gm$ of $S{O}_2$
• C. $11gm$ of dihydrogen
• D. $11gm$ of $S{O}_3$

Answer 1

The correct option is D $11gm$ of $S{O}_3$

For (A)$⟹$ Moles of $C{O}_2=\frac{11}{44}=0.25$

$\therefore$ No. of molecules= $0.25\times N_A$

For (B)$⟹$ Moles of $S{O}_2=\frac{32}{64}=0.5$

$\therefore$ No. of molecules= $0.5\times N_A$

For (C)$⟹$ Moles of $H_2=\frac{11}{2}=5.5$

$\therefore$ No. of molecules= $5.5\times N_A$

For (D)$⟹$ Moles of $S{O}_3=\frac{11}{80}=0.1375$

$\therefore$ No. of molecules=$0.1375\times N_A$

Therefore, $11gms$ of $S{O}_3$ contains less number of molecules.